Question: ${\sqrt[3]{1728} = \text{?}}$
Solution: $\sqrt[3]{1728}$ is the number that, when multiplied by itself three times, equals $1728$ If you can't think of that number, you can break down $1728$ into its prime factorization and look for equal groups of numbers. So the prime factorization of $1728$ is $2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3$ We're looking for $\sqrt[3]{1728}$ , so we want to split the prime factors into three identical groups. Notice that we can rearrange the factors like so: $1728 = 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3 = \left(2\times 2\times 3\right)\times\left(2\times 2\times 3\right)\times\left(2\times 2\times 3\right)$ So $\left(2\times 2\times 3\right)^3 = 12^3 = 1728$ So $\sqrt[3]{1728}$ is $12$.